 Hello, welcome to module 58 of NPTEL NOC, the course on Point Set Apology Part 2. So, we continue with the theme homogeneity. So, today we are reaping the final result of all our efforts last time. So, let x be any n dimensional manifold n greater than or equal to 2. For any k greater than or equal to 1, take two k subsets p 1, p 2, p k, q n, q 2, q k. Take any two k subsets inside x. Given any open connected subset u such that the union of p and q is inside u, the important open connected subset. Contrusion is there exists a homomorphic psi from x to x and an open subset v of v is open inside x such that p union q is inside v contained inside v bar contained inside the original open connected subset u that is the first condition v bar is compact that is the second condition. Psi is identity outside v that is in the complement of v is the third one. The last but least not least important is p i's are mapped to q i for i equal to 1 to 3 to a by psi. So, that is the homogeneity k fold homogeneity. So, let us see how one proves it we have prepared all the requirements already. So, first I will take the case namely p and q are disjoint sets. They are finite sets with k elements p and q are disjoint sets logically this is not necessary but writing down the proof becomes very easy with this approach. You can have different ways of proving this also. So, I choose this one first I choose I assume that p and q are disjoint. Suppose we have completed the proof for this case then first I will show you how to complete the proof in the general case. So, suppose we have proved this case namely p cube empty then what I do is to complete the proof. So, I choose another k subset r r 1 r 2 r k inside u itself which is completely disjoint from p as well as q. Now, I apply the case here namely p and r are disjoint they are k subsets therefore by this assumption I get a homomorphism psi 1 and another psi 2 for q both of them are homomorphisms of x such that there are subsets given v 2 of u satisfying the required conditions. What are the conditions let us just recall p union u p union r is inside v 1 v 1 bar contained inside u similarly for the other case q union r is contained inside v 2 contained inside v 2 bar contained inside u v i bar for compact psi i is identity on the complement of v i i equal to 1 and 2 and psi 1 of p i is r i similarly psi 2 of q i is r i because p and r disjoint and q and r are of a disjoint. Once we have this all that we have to do is take psi equal to p 2 inverse of psi 2 inverse of psi 1 that will map p i to q i p i goes to r i and then the inverse may come back to q i outside v 1 union v 2 both of them will be identity therefore the composition will be also identity v 1 union u 2 being you know closure of that will be v 1 bar union v 2 bar so that is also complex all other things are obvious. Now assume that now p intersection q is empty and prove it then the proof will be completed for this I have to use induction on k when k is 1 we have already proved it in the previous proposition any point can be moved to any other point within a connected open set that we have proved already and for this we do not need even the assumption n greater than equal to 2 even for n equal to 1 we have proved this okay anyway for k greater than equal to 2 we of course need the assumption n greater than equal to 2 we cannot move otherwise because we have already seen that in this example we have seen that for n equal to 1 there are counter examples it is not possible that okay even for r it is not possible to do that always with the full force of this one it is not possible alright so we assume n greater than equal to and make the make this induction hypothesis now okay now where is the induction hypothesis on k minus 1 subsets and on open subsets containing them therefore starting with u which is already given to me and p1p to pk what I do I take the open subset u minus pk qk that is an open subset n greater than equal to 2 therefore it is connected also throwing away finite many points from a connected subset of rn first we have seen for any manifold also connectivity is not destroyed so it is still connected okay so u apply the theorem not for u but u minus this one which contains p1p to pk minus 1 and q1 q2 qk minus 1 okay therefore the induction hypothesis gives you a homeomorphism let us denote website prime and an open subset v prime such that p i q i's are inside v prime now contained inside the v prime bar which is compact the whole thing contained is a u minus pk qk so I apply it to this connected subset such that this psi prime takes p i to q i i equal to 1 2 3 up to k minus 1 and psi prime is identity outside this set v prime in particular it will not disturb pk and qk on pk psi prime of pk pk prime pk pk those things are not disturbed all right so now how to complete the proof once again n greater than 2 I use it follows that u minus all these finite many points 2k minus 1 2k minus 2 in them that is also connected connected open set so it is path connected so we can find a path you look at this one now I am throwing these points so pk and qk are still there inside this one right so because of the path connectivity I can join them by a path omega from 0 1 to u such that omega 0 is pk and omega 1 is qk moreover I do it inside this set now inside this open set that means what this omega does not pass through any of these points 2k minus 2 point okay we can then find a connected open set v double prime of this connected subset such that the image of this path omega 0 1 is contained inside this open subset is connected and its closure is compact contained inside this subset so this is another lemma which we have proved so let me show you this lemma so that was a lemma starting with any path and contained inside an open subset arbitrary open subset actually you can find a path connected open subset u of x such that u bar is compact and a contained inside u contained u bar contained inside v okay you apply that so we have got this thing okay this v double prime here now apply the theorem for just those 2 points that is all get a psi double prime from x to x homeomorphism such that psi prime is identity outside this v double prime and maps pk to qk that's all okay now all that you have to do is take psi as psi double prime composite psi prime psi prime will map pi to qi up to k minus 1 and then psi double prime will take over that one and maps pk to qk whereas psi double prime is identity on all these you know all these points the pi's have gone to qi's but psi double prime does not disturb them okay therefore the psi will be the required map clearly it is compact it is identity outside v prime u and v double prime both of them contained inside so that is a proof so this is a big theorem we have proved now okay thanks to efforts last time we have already done all the preparation so here is a remark recall that if g is acting g is a group acting on a set x we say the action is transitive if even any 2 elements a and b you can find an element g inside the group such that g of a is b any any 2 points can be mapped to each other the same thing has also saying that the orbit space is the entire space x there is single orbit similarly for k greater or equal to 2 we say the g action is k transitive or k fold transitive you can just say k transitive also if for every pair of k subsets a and b of x there exists a one single g such that g of a equal to b that means if we have a 1 a 2 a k here and b 1 b k here g of a i equal to b that also you can manage okay which way you take that that you can you can even demand that now I have just given g a b that is all the above theorem I said that the group this hx namely group of all self-homomorphisms of x okay on a connected and manifold x n greater or equal to 2 is a k fold transitive for all k there was no restriction on k in the previous theorem this fact is very useful in the study of group theoretic properties of hx now what I am going to do is I am going to give you a topological application of this theorem the application itself is useful that I will not be able to do okay so here is the corollary take a connected and manifold okay given any finite subset p of x any finite subset there exists an open subset u of x such that this p is contained inside u that is very easy but the last part is stunning u bar is homeomorphic to dn or just u will be homeomorphic to the open days u bar is homeomorphic to dn is stronger so you think about this one any finite set scattered around the entire manifold so only condition is that x is connected n dimensional manifold okay so any finite set can be contained inside a you know one single disk homeomorphic disk that is the meaning here there is no there is no metric here now we do not we have not fixed any metrics on topological state though they are metering okay so this is an easy corollary now to this theorem let us see how for n greater than or 2 of course first choose any open set v next such that v bar is homeomorphic to dn okay any open set you can choose there are plenty of them okay next choose q elements you know q contained inside v contained inside x where q is a k subset any k subset any finite subset what you want depends upon what e p you have chosen is p is a k subset you choose a k subset here that is all okay inside any open set with which is homeomorphic dn that you can do now the above theorem says that we have homeomorphism p from x to x such that p of p is equal to q all that I have chosen is cardinality of q is equal to cardinality of p that is what you have the k subset okay so p can be mapped to q but what happens now take u equal to p inverse of p v is homeomorphic to dn v bar is homeomorphic to dn this p is defined on the whole of x so p inverse of v bar will be actually equal to u bar that will be homeomorphic to dn as well over and clearly u contains p okay so for n greater than or 2 we have done but I have stated this theorem even for n equal to 1 right from that this theorem earlier this theorem will not be able to give you that for one point there is no need of that theorem so similarly for n equal to 1 you don't need this theorem but something else what is that namely the one which you are going to prove namely we completely classify the manifold one-dimensional manifold connected one-dimensional manifold from that it will be easy because all that you have to do is x equal to either open interval 01 or the circle verify the statement for open interval 01 any finite subset you can take a slightly bigger interval over right similarly for s1 also take a finite subset take any point which is not one of these finite subset if you remove that point rest of them is an open arc which is homeomorphic to open open set so it makes it slightly smaller then its closure will be also homeomorphic to dn over so you just remove a small arc closure arc okay or an open arc which is disjoint from the given finite set so the statement is easy for dimension one case but you should know that we have to do it only for these two cases okay that follows from classification therefore we will not separately prove it all right the classification theorem will be proved so okay now I will give you something about this abstract manifolds okay the following result tells us that after all we could have just took to the study of you know subsets of Euclidean spaces with the all those paraphernalia second countability first darkness etc would have been automatic matrizability would have been automatic so you could have just studied those spaces for manifold why so this is the theorem that I know I am going to state it at least the single result has several implications in topological homotopical and homological properties of a manifold though we shall not be able to entertain them in this course okay so those things will be taken up in the subsequent in algebraic biology course so what is this theorem this is a theorem every n manifold is homeomorphic to a closed subset of R2n plus 1 so that is the meaning of embedding the stronger thing than saying the closed embedding closed subset of R2n plus 1 this is n manifold this is 2n plus 1 so you have to increase the dimension quite a bit namely double plus 1 there is a elaborate statement here take every take any closed random manifold topological manifold the set of embeddings of X inside I2n plus 1 okay that means what they are functions from X2 this I is just 0 cross 1 interval 0 cross 1 raised to n plus 1 you do not have to go even to R whole of R2n plus 1 this is dense in the space to continuous function X2 I2n plus 1 with the compact operand topology so that is the statement of course this is the more elaborate statement than this one take any function continuous function you can approximate it by a embedding in particular there will be an embedding okay you can take a constant function approximate it by embedding means what there will be such a small very small around that point as small as simply there will be a copy of this entire manifold so arbitrary small neighborhoods of every point will contain a copy of the n dimensional manifold so that is the strong conclusion of this theorem okay I will not be able to do this one this is very powerful result I will not be able to do even this one here because these proofs are quite lengthy they do not need many more techniques than what we have done but there are some techniques needed but they are quite lengthy so I will not be able to do that however I will tell you a few things about this namely the proof of this theorem are very lengthy and hard there is a smooth version of this one also which goes under the name easy Whitney embedding theorems for that you have to do differential topology so those proofs are much easier to do also which you may read from many books such as for example my own book however for the topological case there are not so many easy references you are welcome to see the excellent book of food with walman from which I have taken this statement okay so I have given you a correct reference also it is theorem fifth chapter theorem three or you may choose to read a nice proof of embedding theorem from munkres book here we shall be satisfied with an easy proof of a weaker version namely compact case every compact manifold with or without boundary this time I allow it to have boundary also okay that is why I specifically mentioned here is homeomorphic to a close subset of some Euclidean space okay so it can be embedded so I have taken this statement here this one and put extra condition namely it is compact so that we will prove now okay yeah here so being x is being compact right cover x by finitely many open subsets ui 1 less than or y less than okay on each of which there is a homeomorphism fi from ui to a where this a is either Rn or hn see here I am taking ui to a itself this onto so it is homeomorphism onto a onto Rn or onto hn because why I have to put hn because I allow x to be manifold with boundary points also right there may be boundary point so I have to allow this one so there are two different case I have to be allowed as the case may be okay so there only once one issue cover this one by open subset like this x is compact due to the finite cover now go to our standard this is the inverse of stereographic projection eta from Rn to Sn okay Sn minus the north pole to Rn you have the stereographic projection fi whatever eta is the inverse of that okay so this a this this will be the eta of Rn is the entire of Sn minus to north pole that much you know already okay so stereographic projection and then take gi from x to Sn to be the extension of this eta composite fi see fi's are defined only on ui to a what is a a is Rn or hn hn is subset of Rn so there is an eta so eta composite fi makes sense so we get a function from ui to Sn right so gi is an extension of that wherein the rest of the closed set here x minus ui is sent to the north pole you have to see that because of the surjectivity of a this is a continuous function okay so that I will leave to you it is not very difficult so all this gi i equal to 1 2 3 up to k are now continuous functions from x to Sn what is its property on ui it is 1 1 then you are composing with eta so that is also 1 1 so important property of this one is on each ui gi is 1 1 map when you take a 1 1 1 map product with any number of them it will still be 1 1 map so what happens is this g 1 g 2 g cross I have taken because of g 1 this g will be 1 1 on u 1 because of g 2 it will be 1 1 on u 2 and so on right so each of them is 1 1 on u 1 u 2 u 3 they cover it but why the entire thing is 1 1 so that we have to verify so there you have to specifically use how x minus ui has been mapped x minus y goes to north pole none of the points ui goes to north pole okay they are outside Sn minus they are inside Sn minus the north pole so they are not mapped on that's all you have to check okay so check that g itself is a 1 1 map continuous 1 1 this is compact and that is half star therefore it's a homeomorphism on to a closed subset over is it okay next time we shall take up the classification of undamaged manifors